Before embarking on the proof, we recall from the article projective resolution that if G=CnG = C_n is a finite cyclic group of order nn, then there is a projective resolution of ℤ\mathbb{Z} as trivial GG-module:

where the map ℤG→ℤ\mathbb{Z}G \to \mathbb{Z} is induced from the trivial group homomorphism G→1G \to 1 (hence is the map that forms the sum of all coefficients of all group elements), and where DD, NN are multiplication by special elements in ℤG\mathbb{Z}G, also denoted DD, NN:

D≔g−1,D \coloneqq g - 1,

\,

N≔1+g+g2+…+gk−1N \coloneqq 1 + g + g^2 + \ldots + g^{k-1}

The calculations in the proof that follows implicitly refer to this resolution as a means to defining Hn(G,A)H^n(G, A) (in the case A=K*A = K^\ast), by taking cohomology of the induced cochain complex

Proof

Let σ∈ℤG\sigma \in \mathbb{Z}G be an element of the group algebra, and denote the action of σ\sigma on an elementβ∈K\beta \in K by exponential notation βσ\beta^\sigma. The action of the element N∈ℤGN \in \mathbb{Z}G is

which is precisely the normN(β)N(\beta). We are to show that if N(β)=1N(\beta) = 1, then there exists α∈K\alpha \in K such that β=α/g(α)\beta = \alpha/g(\alpha).

By lemma below, the homomorphisms1,g,…,gn−1:K*→K*1, g, \ldots, g^{n-1}: K^\ast \to K^\ast are, when considered as elements in a vector space of KK-valued functions, KK-linearly independent. It follows in particular that

We want to show that if Tr(β)=0Tr(\beta) = 0, then there exists α∈K\alpha \in K such that β=α−g(α)\beta = \alpha - g(\alpha). By the theorem on linear independence of characters (following section), there exists θ\theta such that Tr(θ)≠0Tr(\theta) \neq 0; notice Tr(θ)Tr(\theta) belongs to the ground field kk since g⋅N=Ng \cdot N = N. Put

Proof

where ai∈Ka_i \in K, and assume nn is as small as possible. In particular, no aia_i is equal to 00, and n≥2n \geq 2. Choose g∈Gg \in G such that χ1(g)≠χ2(g)\chi_1(g) \neq \chi_2(g). Then for all h∈Gh \in G we have